1
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

A square shaped aluminium coin weighs $$0.75 \mathrm{~g}$$ and its diagonal measures $$14 \mathrm{~mm}$$. It has equal amounts of positive and negative charges. Suppose those equal charges were concentrated in two charges $$(+Q$$ and $$-Q)$$ that are separated by a distance equal to the side of the coin, the dipole moment of the dipole is

A
$$34.8 \mathrm{~Cm}$$
B
$$3.48 \mathrm{~Cm}$$
C
$$3480 \mathrm{~Cm}$$
D
$$348 \mathrm{~Cm}$$
2
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

If potential (in volt) in a region is expressed as $$\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{xy}-\mathrm{y}+2 \mathrm{yz}$$, the electric field (in $$N C^{-1}$$) at point $$(1,0,1)$$ is

A
$$\mathrm{-7 j}$$
B
$$+7 \mathrm{j}$$
C
$$\mathrm{-6 i+7 j}$$
D
$$\mathrm{6 i-7 j}$$
3
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Five charges, '$$q$$' each are placed at the comers of a regular pentagon of side '$$a$$' as shown in figure. First, charge from '$$A$$' is removed with other charges intact, then charge at '$$A$$' is replaced with an equal opposite charge. The ratio of magnitudes of electric fields at $$\mathrm{O}$$, without charge at $$A$$ and that with equal and opposite charge at $$A$$ is

COMEDK 2024 Morning Shift Physics - Electrostatics Question 5 English

A
4 : 1
B
2 : 1
C
1 : 4
D
1 : 2
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Two charges '$$-q$$' each are fixed, separated by distance '$$2 d$$'. A third charge '$$q$$' of mass '$$m$$' placed at the mid-point is displaced slightly by '$$x$$' $$(x< < d)$$ perpendicular to the line joining the two fixed charges as shown in Fig. The time period of oscillation of '$$q$$' will be

COMEDK 2024 Morning Shift Physics - Electrostatics Question 6 English

A
$$ \mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^2 \mathrm{~d}^3}{\mathrm{q}^2}} $$
B
$$ \mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^3}} $$
C
$$ \mathrm{T}=\sqrt{\frac{4 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}} $$
D
$$ \mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}} $$
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